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dougbell loves solving puzzles at I am proud PuzzleFry member and like my time invested in solving brain teasers.
  • This is a poorly stated problem.  The existing answer makes assumptions that aren’t stated, which result in the conclusion that the information about which child is older is relevant.  It’s not.

    Let’s look at the problem another way.  In the house on the left, we have:

    Accounts / Science
    Boy Boy
    Boy Girl

    In the house on the right we have:

    Younger / Older
    Girl Boy
    Boy Boy

    With the given information, the odds are the same for either house to have a girl.

    Consider a similar scenario with coins.  I flip two pennies and two dimes.  I tell you that the first dime flipped was heads and that the shinier penny is also heads.  In each case I’ve distinguished the two events–it doesn’t matter what I use to distinguish them, it only matters that they have been distinguished.

    In order for the information about which dime was heads to be useful, you have to make an assumption regarding the complete information available to me and *why* I selected to share order information about the dime and not the penny, as well as *why* selected to share information about shininess about the penny.  If I know the result for both dimes, then the information about which was flipped first is useless: I would not revel information about a coin being tails, so I would always select to tell about the dime that came up heads.  Same logic applies to the pennies.  Since the selection bias is to only tell about heads, the information about order or shininess is irrelevant because it is revealed AFTER the selection of which dime to reveal is made.  Information about order, shininess, etc. is only useful if it is ARBITRARILY revealed.

    There is a more relevant piece of information revealed, which is that your friend is “not quite sure” if either family has a girl.  In order for this to be true, two things must be true: your friend does not know the gender of all four children and none of the children your friend does know about is a girl.  This leaves open the possibility that your friend knows the gender of one of the two other children (either the child that likes science or the baby), and that child is a boy.  Given that your friend knows that the other child in the house on the left likes science, I’m going to assume it is more likely that he also knows the gender of the second child in the house on the left, than the gender of the baby.  Based on THIS information, I pick the house on the right.

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  • I know there is already an extensive answer to this question, but I haven’t read it.  Given the length of the existing answer, perhaps my explanation will be simpler.

    There is a 25% chance that all hats are the same color and a 75% chance that there are 2 hats of one color and 1 hat of another color.  The strategy is to have a player pass unless both the hats they see are the same color.  If a player sees two hats of the same color, they respond that their hat is the opposite color.  In this manner, the team always wins when there are multiple hat colors (only the person wearing the hat that is different than the other two responds and correctly identifies the color of their hat), and always loses when all the hats are the same color (they all claim their hat is the wrong color).

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  • There is no paradox here, or anything even close.

    At 99 pounds water and 1 pound solids, the water is 99/(99+1) = 99% by weight of the potatoes.

    Removing 1% of the water (0.99 pounds) makes the ratio of solids be 98.01/(98.01+1) = 98.99% by weight of the potatoes.  (Or, removing water equivalent to 1% of the weight of the potatoes makes the ratio of solids be 98/(98+1) = 98.9899% by weight.)  This amounts to reducing the quantity of water by 1% (or 1 pound), but reducing the ratio of the water by just 0.01% (or 0.010101…%).

    The potential confusion results from linguistics, not math.  It has to do with the two different uses of the word “percentage”.

    Let’s look at another example.  Let’s say that 99% of 100 people surveyed like ice cream, leaving one person (1%) of people that don’t like ice cream.  To reduce the percentage of people that like ice cream to 98%, just 1% (one person) needs to join the group of people that don’t like ice cream, making 2% of people not like ice cream.  However, if instead of an ice cream fan converting I simply eliminate one person that likes ice cream from the survey, instead of 2% of people not liking ice cream, there are just 1.01%.

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  • This is one of the 2400-year-old formulations of Zeno’s paradox, for which there have been a number of explanations over the centuries.

    From a mathematical perspective, the paradox is resolved using a convergent infinite series.  This is a series (such as 1/2 + 1/4 + 1/8 + …) which has an infinite number of terms, the sum of which converges on a finite value (1 in this case).  When you consider that the time it takes to cover each segment converges towards zero along with the distance, you can illustrate that a finite time is required to cover the finite distance, regardless of how finely the distance and time are divided.

    Zeno’s paradox has some interesting consequences when viewed in the context of quantum mechanics.  The entire notion that reality is quantized into discrete increments means that time and distance cannot be infinitely divided into ever smaller chunks.  There is a limit, a smallest distance (Planck length = 1.617 x 10^-35 meters) and a smallest unit of time (Planck time = 5.391 x 10^-44 seconds).  These two units are related: a Planck time is how long it takes a photon moving at the speed of light to travel 1 Planck length.  If time is quantized, then there is no point in time that exists between T and T + 1 Planck time.  Likewise, the photon does not exist at any intermediate position along its path of travel between D and D + 1 Planck length.  At time T it is at point D and at time T + 1 Planck time it is at point D + 1 Planck length.

    In the context of quantum mechanics, Zeno’s paradox is resolved.  The photon does not “move” from D to D + 1 Planck length.  At time T it exists at D and at T + 1 Planck time it exists at D + 1 Planck length.  A distance of 100 m can be repeatedly divided in half about 122 times before it reaches the Planck length and can be divided no more.  The journey starts with a movement of a Planck length rather than the postulated “no distance whatsoever”.

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  • 4

    Each column sums to 14:

    7 + 2 + 1 + 4 = 14
    3 + 8 + 1 + 2 = 14
    6 + 5 + 2 +1 = 14
    2 + 4 + 4 + 4 = 14

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  • Genius Asked on 22nd October 2015 in Puzzles.


    The solution can be determined by only looking at the circle with the missing number.  Starting with the lowest number (1), move clockwise incrementing by 4:

    1, 5, 9, *13*, 17, 21

    Arguably, the answer could be determined with question marks for every slice except one in the circle since the upper left increments by 2, the upper right by 3 and the lower left by 5: starting from the upper left circle, the increment increases by 1 as you move clockwise to the next circle and the minimum value is always in the same position.

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  • 8, 1

    The phrasing of the question (number instead of numbers) provides the clue that each pair of digits in a row forms a two-digit number.  So representing the grid as two-digit numbers we have:

    73, 46, 19  (delta -27)
    11, 9, 7  (delta -2)
    52, 42, 32  (delta -10)
    99, 50, 1  (delta -49)
    67, 82, 97  (delta 15)
    15, 48, 81  (delta 33)

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  • 6

    There are two ways to justify 6 as the answer:

    1) 6 is the missing value between 1 and 9
    2) 6 makes each row and column sum to 15 (aka a magic square)

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